Question

1) Develop an argument on the fact that Even parity (4,3) code is a linear block code whereas an Odd parity (4,3) is not based on the three basic conditions for linear block codes.

2) What is hamming distance? Write Cyclic codes of a (4,7) encoding scheme and calculate the minimum and maximum hamming distance.

3) Draw the look-up table, trellis diagram, tree diagram and state diagram for a (2,1,4) convolution encoding scheme.

Answer 1

Part 1

All columns of H have odd parity, so the sum (exclusive-or) of an odd number of columns has odd parity and is therefore nonzero. Every code word selects a set of columns of H whose sum is zero, so code words must have an even number of nonzero components.

BLOCK CODE LINEAR The first k bits of a (n,k) linear block coding are always identical to the message sequence to be sent. The second component of (n-k ) bits is calculated from message bits using the encoding rule and is known as parity bits.

Part 2

The Hamming distance between two strings of similar length in information theory is the number of locations where the corresponding symbols differ.

The (7,4) Hamming code has a generator polynomial ${\displaystyle g(x)=x^{3}+x+1}$ . This polynomial has a zero in Galois extension field ${\displaystyle GF(8)}$  at the primitive element ${\displaystyle \alpha }$ , and all codewords satisfy ${\displaystyle {\mathcal {C}}(\alpha )=0}$ ${\displaystyle GF(2)}$ . Blocklength will be ${\displaystyle n} equal to {\displaystyle 2^{m}-1}$  and primitive elements ${\displaystyle \alpha } and {\displaystyle \alpha ^{3}}$  as zeros in the ${\displaystyle GF(2^{m})}$  because we are considering the case of two errors here, so each will represent one error.

The received word is a polynomial of degree ${\displaystyle n-1}$  given as ${\displaystyle v(x)=a(x)g(x)+e(x)}$

where ${\displaystyle e(x)}$  can have at most two nonzero coefficients corresponding to 2 errors.

We define the Syndrome Polynomial, ${\displaystyle S(x)}$  as the remainder of polynomial ${\displaystyle v(x)}$  when divided by the generator polynomial ${\displaystyle g(x)} i.e. {\displaystyle S(x)\equiv v(x)\equiv (a(x)g(x)+e(x))\equiv e(x)\mod g(x)} as\\ {\displaystyle (a(x)g(x))\equiv 0\mod g(x)}.$

Part 3